Proof. Suppose that $X,Y\in M_n(R)$ satisfy $XY=I$ . Since $R/J$ is stably finite then $I-YX\in M_n(J)$ . Since J is a quasi-invertible ideal, so is $M_n(J)\triangleleft M_n(R)$ and hence $YX=I-(I-YX)$ is invertible, so both X, Y are invertible and since $XY=I$ it follows that $X=Y^{-1}$ and $YX=I$ .

Proof. Suppose that R is a PI-ring and $J\triangleleft R$ is its Jacobson radical. Each primitive homomorphic image of R is a simple algebra which is finite-dimensional over its center [ Reference Kaplansky9 , theorem 1], hence stably finite by linear algebra. Thus, $R/J$ embeds into a direct product of stably finite rings, so it is stably finite itself. By Lemma 2·1, R itself is stably finite.

Proof. On the contrary, if A is linear sofic then we have an embedding:

\begin{align*} \Phi\;:\; A \longrightarrow \prod_{k\rightarrow \mathcal{U}} M_{n_k}(F) / \mathrm{Ker}(\rho_\mathcal{U}). \end{align*}

Fix a linear lift of $\Phi$ to $\prod_k M_{n_k}(F)$ , say, $\widehat{\varphi}\;:\; A \rightarrow \prod_k M_{n_k}(F)$ so:

\begin{align*} \Phi(a)=0 \iff \widehat{\varphi}(a) \in \mathrm{Ker}(\rho_\mathcal{U}). \end{align*}

Write $\widehat{\varphi}=\prod_k \varphi_k$ with each $\varphi_k\;:\; A\rightarrow M_{n_k}(F)$ . For every $0\neq a\in A$ there exists $\varepsilon>0$ such that:

\begin{align*} \{k\;:\; \mathrm{rk}(\varphi_k(a))>\varepsilon n_k\} \in \mathcal{U}, \end{align*}

and for every $a,b\in A$ and $\varepsilon>0$ , we have:

\begin{align*} \{k\;:\; \mathrm{rk}\!\left(\varphi_k(a)\varphi_k(b)-\varphi_k(ab)\right)<\varepsilon n_k\}\in \mathcal{U}. \end{align*}

Since $z\neq 0$ , it follows that also $x\neq 0$ . In particular, since $\mathcal{U}$ is a non-principal ultrafilter, we can fix a positive real $\alpha>0$ and a linear map $\varphi=\varphi_k\;:\; A\rightarrow M_n(F)$ for some k such that:

\begin{align*} \mathrm{rk}(\varphi(x)),\ \mathrm{rk}(\varphi(z)) > \alpha n. \end{align*}

By the assumptions of the lemma, $x=yxa$ and $z=xb$ for some $a,b\in A$ , and $yz=0$ . Let $T\;:\!=\;\varphi(x)-\varphi(y)\varphi(x)\varphi(a)$ and $S\;:\!=\;\varphi(z)-\varphi(x)\varphi(b)$ . We may additionally assume that:

\begin{align*} \mathrm{rk}(\varphi(y)\varphi(z)),\ \mathrm{rk}(T),\ \mathrm{rk}(S) < \frac{\alpha}{4}n. \end{align*}

Claim. We have:

\begin{align*}\mathrm{rk}(\varphi(y)\varphi(x)) < \mathrm{rk}(\varphi(x))-\frac{\alpha}{2} n.\end{align*}

Proof of claim. First, since $\varphi(z)=\varphi(x)\varphi(b)+S$ :

\begin{align*} \mathrm{Im}(\varphi(z))\subseteq \mathrm{Im}(\varphi(x)) + \mathrm{Im}(S). \end{align*}

It follows that:

\begin{eqnarray*}\frac{\mathrm{Im}(\varphi(z))}{\mathrm{Im}(\varphi(z)) \cap \mathrm{Im}(\varphi(x))} & \cong & \frac{\mathrm{Im}(\varphi(z))+\mathrm{Im}(\varphi(x))}{\mathrm{Im}(\varphi(x))} \\[5pt] & \subseteq & \frac{\mathrm{Im}(\varphi(x))+\mathrm{Im}(S)}{\mathrm{Im}(\varphi(x))} \\[5pt] & \cong & \frac{\mathrm{Im}(S)}{\mathrm{Im}(\varphi(x)) \cap \mathrm{Im}(S)}.\end{eqnarray*}

So:

\begin{align*} \dim_F \left( \mathrm{Im}(\varphi(z))\cap \mathrm{Im}(\varphi(x))\right) \geqslant \mathrm{rk}(\varphi(z))-\mathrm{rk}(S)>\frac{3\alpha}{4} n. \end{align*}

Denote $\mathcal{V} \;:\!=\; \mathrm{Im}(\varphi(y)\varphi(z))$ and recall that $\dim_F \mathcal{V} < {\alpha}n/{4}$ . Fix a direct sum complement of $\mathrm{Im}(\varphi(z))\cap \mathrm{Im}(\varphi(x))$ inside $\mathrm{Im}(\varphi(x))$ , say, $\mathcal{W}$ , and notice that:

\begin{eqnarray*} \dim_F \mathcal{W} & = & \mathrm{rk}(\varphi(x)) - \dim_F \left( \mathrm{Im}(\varphi(z))\cap \mathrm{Im}(\varphi(x)) \right) \\[5pt] & < & \mathrm{rk}(\varphi(x)) - \frac{3\alpha}{4} n. \end{eqnarray*}

Now:

\begin{eqnarray*}\mathrm{rk}(\varphi(y)\varphi(x)) & = & \dim_F \mathrm{Im}(\varphi(y)\varphi(x)) \\[5pt] & = & \dim_F \varphi(y)\mathrm{Im}(\varphi(x)) \\[5pt] & \leqslant & \dim_F \varphi(y)\left(\mathrm{Im}(\varphi(z)) \cap \mathrm{Im}(\varphi(x)) \right) + \dim_F \varphi(y)\mathcal{W} \\[5pt] & \leqslant & \dim_F \mathcal{V} + \dim_F \mathcal{W} \\[5pt] & < & \frac{\alpha}{4}n + \left(\mathrm{rk}(\varphi(x)) - \frac{3\alpha}{4} n \right) \\[5pt] & = & \mathrm{rk}(\varphi(x))-\frac{\alpha}{2} n.\end{eqnarray*}

Return to the proof of the lemma. Since $\varphi(x)=\varphi(y)\varphi(x)\varphi(a)+\;T$ for some $T\in M_n(F)$ with $\mathrm{rk}(T)<{\alpha}n/{4}$ , we have:

\begin{eqnarray*}\mathrm{Im}\;\varphi(x) & = & \mathrm{Im}(\varphi(y)\varphi(x)\varphi(a)+T) \\[5pt] & \subseteq & \mathrm{Im}\!\left( \varphi(y) \varphi(x)\right) + \mathrm{Im}(T)\end{eqnarray*}

whose dimension is at most $ \mathrm{rk}\!\left(\varphi(y)\varphi(x)\right) + \mathrm{rk}(T) $ which is, by the above claim, at most:

\begin{align*} \mathrm{rk}(\varphi(x))-\frac{\alpha}{2} n + \mathrm{rk}(T) < \mathrm{rk}(\varphi(x))-\frac{\alpha}{4} n, \end{align*}

a contradiction. Hence A is non linear sofic.

Proof of Theorem 1·2. Let $A=F\langle x,y \rangle / \langle x^2, yxy-x \rangle $ . This algebra was introduced by Irving [ Reference Irving6 ] as an example of a finitely presented PI algebra which is not embeddable into any matrix algebra over a field.

The set of monomials in x, y which avoid occurrences of $x^2$ and yxy forms a linear basis for A; this fact was established in [ Reference Irving6 ]. Indeed, this is a direct consequence of Bergman’s Diamond Lemma [ Reference Bergman2 ], since the only overlap between the reductions $yxy\mapsto x,\ x^2\mapsto 0$ is:

\begin{align*} 0=xxy=(yxy)xy=yx(yxy)=yxx=0. \end{align*}

By [ Reference Irving6 , theorem 2], A satisfies a polynomial identity (and has linear growth). Explicitly, since $\langle x\rangle \triangleleft A$ satisfies $A/\langle x\rangle \cong F[y]$ and $\langle x\rangle ^3=0$ , the identity:

\begin{align*} [X_1,Y_1][X_2,Y_2][X_3,Y_3]=0 \end{align*}

holds in A. In particular, by Lemma 2·2, A is stably finite.

Finally, A fulfills the requirements of Lemma 3·1 with $z=xyx$ . Indeed, xyx contains no occurrences of $x^2$ or yxy, and is thus non-zero; obviously, $xyx\in xA$ and $x=yxy\in yxA$ ; and finally,

\begin{align*} yz=yxyx=x^2=0. \end{align*}

Hence A is non linear sofic.

Proof of Theorem 1·3. Let $A=F\langle \langle x,y\rangle \rangle $ be the ring of noncommutative formal power series and let $A^{+}$ be the ideal of A consisting of all power series with zero constant term. Let $I=\langle yx^2y-x\rangle \triangleleft A^{+}$ . Since $A^{+}$ is Jacobson radical, the quotient ring $R\;:\!=\;A^{+}/I$ is also Jacobson radical. By [ Reference Cohn and Sa̧siada3 , section 2, “basic lemma”], the image of x in R (for simplicity, we identify elements in $A^{+}$ with their images modulo I, by abuse of notation) is non-zero. Consider the ring:

\begin{align*} S = \left( \begin{matrix}R & R \\[5pt] 0 & 0\end{matrix} \right) \end{align*}

and consider the ideal:

\begin{align*} K\;:\!=\; \left\langle \left( \begin{matrix}0 & y x^2 \\[5pt] 0 & 0\end{matrix} \right)\right\rangle \triangleleft S. \end{align*}

We claim that:

\begin{align*} (\!*\!)\ \ \ \left( \begin{matrix}0 & x^2 \\[5pt] 0 & 0\end{matrix} \right) \notin K. \end{align*}

Indeed, a straightforward calculation shows that $KS=0$ , so if $(\!*\!)$ was not true then:

\begin{eqnarray*} \left( \begin{matrix}0 & x^2 \\[5pt] 0 & 0\end{matrix} \right) & = & \alpha \cdot \left( \begin{matrix}0 & y x^2 \\[5pt] 0 & 0\end{matrix} \right) + \sum_{i=1}^{n} \left( \begin{matrix}a_i & b_i \\[5pt] 0 & 0\end{matrix} \right) \left( \begin{matrix}0 & y x^2 \\[5pt] 0 & 0\end{matrix} \right) \\[5pt] & = & \left( \begin{matrix}0 & \alpha y x^2 + \sum_{i=1}^{n} a_i y x^2 \\[5pt] 0 & 0\end{matrix} \right),\end{eqnarray*}

(where $\alpha\in F$ is some – possibly zero – scalar.) Considering the upper right corner of the above equation, we obtain $x^2=fx^2$ where $f=\alpha y+\sum_{i=1}^{n} a_iy\in R$ . Let g be the quasi-inverse of f, namely, $gf=f+g$ . Then $gx^2=gfx^2=fx^2+gx^2$ , so $x^2=fx^2=0$ , hence $x=yx^2y=0$ . This contradicts that $x\neq 0$ . It follows that $(\!*\!)$ holds.

Consider $T\;:\!=\;S/K$ and consider the following elements, identified with their images modulo K:

\begin{align*} X\;:\!=\;\left( \begin{matrix}x & 0 \\[5pt] 0 & 0\end{matrix} \right),\ Y\;:\!=\;\left( \begin{matrix}y & 0 \\[5pt] 0 & 0\end{matrix} \right),\ Z\;:\!=\;\left( \begin{matrix}0 & x^2 \\[5pt] 0 & 0\end{matrix} \right) \end{align*}

Notice that in T, it holds that $X=YX^2Y\in YX\cdot T$ and:

\begin{align*} Z=\left( \begin{matrix}x & 0 \\[5pt] 0 & 0\end{matrix} \right)\left( \begin{matrix}0 & x \\[5pt] 0 & 0\end{matrix} \right)\in X\cdot T \end{align*}

and in addition:

\begin{align*} YZ=\left( \begin{matrix}y & 0 \\[5pt] 0 & 0\end{matrix} \right)\left( \begin{matrix}0 & x^2 \\[5pt] 0 & 0\end{matrix} \right)=\left( \begin{matrix}0 & yx^2 \\[5pt] 0 & 0\end{matrix} \right) \end{align*}

which belongs to K, hence equal to 0 in T. Moreover, by $(\!*\!)$ it follows that Z is non-zero in T.

By Lemma 3·1 applied to T with respect to X, Y, Z, the algebra T is non linear sofic; since S is Jacobson radical, so is $T=S/K$ . The unital hull $T^{1}$ is non linear sofic and stably finite by Lemma 2·1.